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3 years ago

The distance between stars is typically measured in

Physics

2 answers:

Gelneren [198K]3 years ago

7 0

The correct answer is Light years

igor_vitrenko [27]3 years ago

6 0

Answer:

Light year

Explanation:

The distance between the stars is very large. The unit measure this distance should be very large. It is called light year.

One light year is defined as the distance traveled by the light in one year.

Speed of light is 3 x 10^8 m/s

Time taken is 1 year = 365 days = 365 x 24 x 60 x 60 = 31536000 second

By using the formula

distance = speed x time

distance = 3 x 10^8 x 31536000 = 9.46 x 10^15 m

Thus , one light year is equal to 9.46 x 10^15 m.

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The two waves shown here are the same distance from point x and are traveling toward each other

kompoz [17]

The Answer should be A: 0.0

3 0

2 years ago

Read 2 more answers

an object with a mass of 7.5 kg accelerates 8.3 m/s2 when an unknown force is applied to it. what is the amount of the force?

nasty-shy [4]

F=MA
F is force
M is mass
A is acceleration
F=MA
F=(7.5 kg)(8.3 m/s^2)
F=62.25 N (newton- SI unit for force)
Hope this helps

5 0

3 years ago

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos

deff fn [24]

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\$

Electric field by charge $q_1$ at point A,

$F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C$